Direct Coulomb energy in the spherical symmetry

The integral for the direct Coulomb energy in spherical symmetry has a integrand which has a discontinuous first derivative. In order to get an integral which is easier to calculate using Gauss-Hermite integration we use the 'Vautherin trick' [8]

$$\frac{e^2}{2} \int{\rm d}^3\vec{r}\,{\rm d}^3\vec{r}'\, \rho\left(r\right)\frac{1}{\left\vert\vec{r}-\vec{r}'\right\vert}\rho\left(r'\right)$$

$$= \frac{\left(4\pi\right)^{2}e^2}{6}\int{\rm d}\,{r}\,{\rm d}\,{r}'... ...\vert r-r'\right\vert^{3}\right]rr'\rho\left(r\right)\Delta'\rho\left(r'\right)$$

$$= \frac{4\pi e^2}{2\sqrt{3}}\int {\rm d}\,{r}\,{\rm d}\,{r}'\, \tilde{\rho}_{00,0000,0}\left(r\right)re^{-(br)^{2}}$$

$$\times \int {\rm d}\,{r}\, \left[\left(r+r'\right)^{3}-\left\ver... ...r'\right\vert^{3}\right]r'\tilde{\rho}_{20,0000,0}\left(r'\right)e^{-(br')^{2}}$$

$$= \frac{4\pi e^2}{2 \sqrt{3}}\int\tilde{\rho}_{00,0000,0}\left(r\right)re^{-(br)^{2}}V_{DC}\left(r\right)$$ (117)

which gives a smoother integrand. However in order to perform the linear-response calculations, where expressions for non-spherical multipoles are needed, we also developed a different method which will be presented elsewhere [9].